I'm not sure if my question makes sense, but I don't know where to start.

A crate of mass $m_2$ is connected by a massless rope of constant length $l$ to a rocket of mass $m_1$. We take into account gravity (the crate is initially hanging directly underneath), but the rocket has no vertical acceleration since it's acted upon by an upward force of $(m_1+m_2)*g$. We neglect air drag and treat the rocket and the crate it is towing as blocks of the same size. If a horizontal force $F$ acts upon the rocket, what happens to the crate? It's probably going to lag behind at a certain angle, but how can we find that angle? Is the situation different if there's no gravity?

Edit: This link helped me answer my question: http://cnx.org/content/m14061/latest/?collection=col10322/1.83


When you have a problem like this the first thing to do is draw a diagram. Mark in all the angles and all the forces you think are acting. I think the diagram looks like this:

Rocket towing mass

Your question is what angle, $\theta$, does the rope make to the vertical. Since this is a homework problem I'm not going to answer it directly, but note that the horizontal acceleration of the mass, $m_2$, must be equal to the acceleration of the rocket otherwise the mass would be accelerating relative to the rocket. A bit of basic trigonometry and you should have your angle.

  • $\begingroup$ The acceleration of both should be same and not forces. Same forces -> different accelerations. $\endgroup$ – Rijul Gupta Jun 7 '14 at 6:26
  • $\begingroup$ @JohnRennie, what did you use to create that picture, thanks $\endgroup$ – Derg Jun 7 '14 at 7:42
  • $\begingroup$ @Derg: Google Draw. It's a fairly basic drawing app but it's free and available everywhere. I do the drawing then screen shot it and save it as a GIF. $\endgroup$ – John Rennie Jun 7 '14 at 7:46

Assuming no air-drag/air-friction following two cases will be observed

Take note, you can get wrong answers while working with forces, try to work with accelerations here.

In presence of gravity

enter image description here

Here you can easily find $\theta$ using basic trigonometry

$\tan\theta = \frac{g}{a}$

In absence of gravity

enter image description here

Clearly the angle here is $0^\text{o}$

  • $\begingroup$ How do we justify the assumption that the horizontal distance between the two objects is equal to $a$ (in the presence of gravity)? $\endgroup$ – Simeon Jun 7 '14 at 10:27
  • $\begingroup$ That's the acceleration. Please see again "a" is written in front of a ray and not on a line segment. $\endgroup$ – Rijul Gupta Jun 7 '14 at 10:47
  • $\begingroup$ Yes, but if we're going to use trigonometry, we'll have to have to treat $a$ as a length, won't we? $\endgroup$ – Simeon Jun 7 '14 at 10:50
  • $\begingroup$ Have you never used velocties (vertical and horizontal) in a parabola to find angle of trajectory? No! We do not have to treat it as length. $\endgroup$ – Rijul Gupta Jun 7 '14 at 10:59

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